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A solution with $\mathrm{pH}=4$ is mixed with a solution with $\mathrm{pH}=10$ in equal volumes. Calculate resultant $\mathrm{pH}$. Please note both solution consists of strong acids and/or bases i.e. They will completely dissociate.

$Method 1:$

Let volume be 1L of each

pH of solution 1=4

Concentration of H is $10^{-4} $

pH of solution 2=10

Concentration of H is $10^{-10} $

Resultant concentration =$\frac{10^{-4} +10^{-10} }{2}$

Therefore calculting pH we get 4.301

$Method 2:$

pH of solution 1=4

Concentration of H is $10^{-4} $

pH of solution 2=10

pOH=14-10=4

Concentration of OH=$10^{-4} $

Hence the H and OH neutralizes each other so it will result in neutral solution ph=7

Edit: I found another method which is kind of same as method 2

Method 3

In Solution 1: pH=4. Concentration of H is $10^{-4} $

Since pOH=14-pH. Hence pOH=10. Therefore concentration of OH=$10^{-10} $

In solution 2

pH=10. Concentration of H is $10^{-10} $

Since pOH=14-pH. Hence pOH=4. Therefore concentration of OH=$10^{-4} $

So the total concentration of H=concentration of OH

Hence neutrilization reaction would occur and resultant pH will be 7

Which method is right and why??

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  • $\begingroup$ The second method, so to put it, is slightly less wrong. But in fact we can't calculate the resulting pH at all. We don't have the necessary data. Knowing the pH of original solutions is simply not enough. $\endgroup$ – Ivan Neretin May 27 at 13:31
  • $\begingroup$ Can you please explain why is it so?? $\endgroup$ – Satwik May 27 at 13:32
  • $\begingroup$ What else is required?? $\endgroup$ – Satwik May 27 at 13:33
  • $\begingroup$ We need to know what compounds are there. $\endgroup$ – Ivan Neretin May 27 at 13:35
  • $\begingroup$ Why do we need that?? I couldn't get your point. I think if it is given that solution2 has pH 10 it will be basic and that means we have a basic compound and the other will be acidic as its pH is 4 $\endgroup$ – Satwik May 27 at 13:39

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